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Answer:
The mass flow rate of refrigerant is 0.352 kg/s
Explanation:
Considering the cycle of an ideal heat pump, provided in the attachment, we first find enthalpy at state B and D. For that purpose, we use property tables of refrigerant R134a:
<u>At State A</u>:
From table, we see the enthalpy and entropy value of saturated vapor at 0.2 MPa. Therefore:
ha = 244.5 KJ/kg
Sa = 0.93788 KJ/kg.k
<u>At State B</u>:
Since, the process from state A to B is isentropic. Therefore,
Sb = Sa = 0.93788 KJ/Kg
From table, we see the enthalpy value of super heated vapor at 1 MPa and Sb. Therefore:
hb = 256.85 KJ/kg (By interpolation)
<u>At State C</u>:
From table, we see the enthalpy and entropy value of saturated liquid at 1 MPa. Therefore:
hc = 107.34 KJ/kg
Now, from the diagram it is very clear that:
Heat Loss = m(hb = hc)
m = (Heat Loss)/(hb - hc)
where,
m = mass flow rate = ?
Heat Loss = (180,000 Btu/hr)(1.05506 KJ/1 Btu)(1 hr/3600 sec)
Heat Loss = 52.753 KW
Therefore,
m = (52.753 KJ/s)/(256.85 KJ/kg - 107.34 KJ/kg)
<u>m = 0.352 kg/s</u>
Answer:
power developed by the turbine = 6927.415 kW
Explanation:
given data
pressure = 4 MPa
specific enthalpy h1 = 3015.4 kJ/kg
velocity v1 = 10 m/s
pressure = 0.07 MPa
specific enthalpy h2 = 2431.7 kJ/kg
velocity v2 = 90 m/s
mass flow rate = 11.95 kg/s
solution
we apply here thermodynamic equation that
energy equation that is
put here value with
turbine is insulated so q = 0
so here
solve we get
w = 579700 J/kg = 579.7 kJ/kg
and
W = mass flow rate × w
W = 11.95 × 579.7
W = 6927.415 kW
power developed by the turbine = 6927.415 kW